Center of Mass, Momentum & Collision

JEE Main level, single correct, medium

101 of 130

Question 101 of 130

A rubber ball is released from a height of 5m5m above the floor. It bounces back repeatedly, always rising to 81100\dfrac{81}{100} of the height through which it falls. Find the average speed of the ball. (Take g=10ms−2g=10{ms}^{-2})

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Answer: (D) 2.50ms−12.50{ms}^{-1}

v0=2gh{v}_{0}=\sqrt{2gh}
v=e2gh=2ghv=e\sqrt{2gh}=\sqrt{2gh}
⇒e=0.9\Rightarrow e=0.9
S=h+2e2h+2e4h+………S=h+2{e}^{2}h+2{e}^{4}h+\ldots \ldots \ldots
t=2hg+2e2hg+2e22hg+………t=\sqrt{\dfrac{2h}{g}}+2e\sqrt{\dfrac{2h}{g}}+2{e}^{2}\sqrt{\dfrac{2h}{g}}+\ldots \ldots \ldots
vav=st=2.5ms−1{v}_{av}=\dfrac{s}{t}=2.5m{s}^{-1}

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